What volume of nitrogen dioxide is produced from the complete reaction of 16.87g of lead at STP?

2Pb (NO3) - -> 2PbO (s) + 4NO2 (g) + O2 (g)

Answer: 16.78 g of lead (II) nitrate will produce 2.238 L of NO2

Explanation:

From the balanced equation, it is clear that every mole of lead nitrate will produce 2 moles of NO2. n (no. of moles) = mass/molar mass

n (lead nitrate) = 16.87/331.2 = 0.05 mole

0.05 mole of lead nitrate will produce 0.1 mole of nitrogen dioxide (NO2)

PV = nRT the general gas law

at STP: P = 1.0 atm and T = 273.0 K

R = 0.082 L. atm/mol. K

V = nRT/P

V = (0.10 mol) (0.082 L. atm/mol. K) (273.0 K) / (1.0 atm) = 2.238 L